He mentioned additive inverses and closure under addition (implying 0 being an element), so he has no problem subtracting things. The key lies in the one part of the definition of a vector space that is left out.

He mentioned additive inverses and closure under addition (implying 0 being an element), so he has no problem subtracting things. The key lies in the one part of the definition of a vector space that is left out.

I misread it. I thought the question asked to find something that fails to be a subspace because it's not closed under additive inverses.

first try to solve a simpler problem.
are there nontrivial subgroups of the group [tex] (\mathbb{R},+,-,0) [/tex] ? that means is there a set [tex] A [/tex] with [tex] \{0\} \subset A \subset \mathbb{R} [/tex] such that [tex] A [/tex] is closed under addition and substraction?
once you have found such a subgroup [tex] A [/tex], is [tex] A^2 [/tex] a set with the desired property?

In order that a subset be a subspace, it must be closed under addition, additive inverses, and scalar multiplication. Since your subset is required to be closed under addition and additive inverse, there's only one place left to look!